Determine the minimum concentration of the precipitating agent on the right to cause precipitation of the cation from the solution on the left. c. 0.0018 M AgNO₃; RbCl

Video transcript

Hello everyone today we are being asked to calculate the smallest concentration of potassium bromide required to precipitate the Catalan in a . moller lead nitrate solution. And so the first thing you want to do is you want to draw out what these process what these reactions would look like. So if we reacted potassium bromide in an agreed solution with lead nitrate. Also Equus we would get lead bromide as a solid and potassium nitrate in the acquis form. Now this isn't balanced but too easily balance this. We can simply add a two to the left and a two in front of the potassium nitrate and it's balanced. We then take the solid that we formed and its own equation. So we say that led br two solid is going to form lead two plus Aquarius as well as to be r minus in the acquis form and the K. S. P. I'm gonna write on the site here of this of this lead bromide is 4.67 times 10 to the negative six. R. K. S. P. Equation is going to be K. S. P. Is equal to our products which has led to plus times our B. R minus. And since we have a coefficient of two, that's going to be the exponent for bromide. And so once again we're going to say that we have lead nitrate and the acquis forming lead two plus in the acquis form as well as two nitrate in the acquis form. And so we can notice that there's a 1-1 ratio with our lead and our lead nitrate. And so therefore our concentration of lead two plus is going to be 0.018 moller plugging this into our K. Sp expression. We say that the K. S. P, which is 4.67 times 10 to the negative six is equal to the concentration of lead two plus, which is 0.18 is equal to B. R squared. Which was we're going to just say is X squared Simplifying our concentration. We can see the concentration of X is equal to 0. Molar. And so finally we're going to show that when Kbr and the acquis dissociates, we're going to form K plus in the acquis ions as well as B r minus in the acquis form. And we see that our concentration of B r minus is equal to our concentration of Kbr Which means that we're ending up with a final concentration of 0. molar. I hope this helped. And until next time